Skip to content

Sunday Times Teaser 3013 – Arian Pen-Blwydd

by Howard Williams

Published Sunday June 21 2020 (link)

When I thought that my daughter was old enough to be responsible with money I gave her on her next, and all subsequent birthdays, cash amounts (in pounds) which were equal to her birthday age squared.

On her last birthday her age was twice the number of years for which she received no such presents. I calculated at this birthday that if I had made these gifts on all of her birthdays then she would have received 15% more than she had actually received. I then decided that I would stop making the payments after her birthday when she would have received only 7.5% more if the payments had been made on all of her birthdays.

What was the amount of the final birthday payment?

3 Comments Leave one →
  1. Brian Gladman permalink

    Let \(S(n)\) be the sum of the perfect squares from 1 to \(n\), which is \(n(n+1)(2n+1)/6\). Let \(m+1\) and \(n\) be the first and last paid birthdays. The first condition is

    \[S(2m)=(115/100)\left\{S(2m)-S(m)\right\}\Rightarrow 3\,S(2m)=23\,0S(m)\]

    On expanding the sums of squares, this becomes

    \[3\,(2m)(2m+1)(4m+1)/6=23\,m(m+1)(2m+1)/6\]which can be simplified to give:\[m(2m+1)(m-17) = 0\]Since we are only interested in positive values for m, this gives \(m = 17\).

    The second condition is:\[S(n)=(215/200)\left\{S(n)-S(m)\right\}\Rightarrow 3\,S(n)=43\,S(m)\]

    Expanding and substituting for \(m\) now gives \[n(n+1)(2n+1)=2\times3\times5\times7\times17\times43\] Here \(2n^3<153510\), which means that \(n < 43\), and one of the terms on the left hand side must have 43 as a factor since 43 on the right hand side is a prime. Hence either \((n + 1) = 43\) or \((2n+1)=43\); substituting these values into the equation shows that only \(n=42\) provides a solution so the above equation simplifies to: \[(n-42)(2\,n^2+ 87\,n+3655)\;=\;0\]where the quadratic term might give two additional solutions. In fact it doesn't, since it has two complex conjugate roots: \(n=(1/4)(-87\pm\sqrt{21671}\,i)\) and this leaves \(n=42\) as the only real solution. The answer is hence a final payment of £1764.

  2. Tony Smith permalink


    I strongly approve of your efforts to (a) prove the uniqueness of solutions and (b) correct false statements when they appear.

    It has effectively been claimed (without proof) on your other site that
    (m, n) = (17, 42) is the only solution in positive integers of the Diophantine equation
    43m(m+1)(2m+1) = 3n(n+1)(2n+1).

    I would like to know whether the claim is true.


    • Brian Gladman permalink

      Hi Tony, Its good to know that you share my thoughts on such matters!

      Turning to cubic diophantine equations, these are notoriously difficult to solve but there has been enormous progress in recent years with elliptic curves. This paper looks promising since I believe our cubic can be converted into a solvable form and we already have one rational point to kick off the analysis. But it is hard work and I am not sure that I will have the time that understanding this paper demands.

Leave a Reply

Note: HTML is allowed. Your email address will not be published.

Subscribe to this comment feed via RSS